Answer
424.5k+ views
Hint: We solve the given question by noticing the fact that distance between any two points on the cube is maximum when they are the endpoints of the diagonal of the cube. So, we start by finding the length of the diagonal of any face using Pythagoras theorem and then we select another side of the cube such that they form a right-angled triangle and use the Pythagoras theorem again to find the length of the diagonal.
Complete step-by-step answer:
We were given that the length of the side of the cube is 10cm.
We need to find the length of the longest rod that has to be fitted in the cube. As the rod has to touch the cube at two points, it means we have to find the longest distance between any two points on the sides of the cube.
Longest distance between two points of the cube is when it is the diagonal of the cube.
Let the length of the side of the cube be $ a=10cm $ .
The longest distance is the diagonal of the cube which is AG from the above figure.
First, let us find the length of the side AF which is the diagonal of the face ABFE.
As ABFE is the face of a cube, it is square with side length $ a=10cm $ . Then $ \Delta ABF $ is a right-angled triangle.
So, by using the Pythagoras theorem, which is
If a, b are adjacent sides of a right-angled triangle and c is the hypotenuse, then
$ {{a}^{2}}+{{b}^{2}}={{c}^{2}} $ .
So, by applying the Pythagoras Theorem on the $ \Delta ABF $ , we get
$ \begin{align}
& \Rightarrow \text{ }{{\left( AB \right)}^{2}}+{{\left( BF \right)}^{2}}={{\left( AF \right)}^{2}} \\
& \Rightarrow \text{ }{{\left( a \right)}^{2}}+{{\left( a \right)}^{2}}={{\left( AF \right)}^{2}} \\
& \Rightarrow \text{ }{{\left( AF \right)}^{2}}=2{{a}^{2}} \\
& \Rightarrow \text{ }AF=\sqrt{2}a \\
\end{align} $
Now consider the $ \Delta AFG $ , where $ AF=\sqrt{2}a $ , $ FG=a $ as it is a side of the cube and $ \angle F={{90}^{\circ }} $ .
So, by applying the Pythagoras Theorem again on the $ \Delta ABF $ , we get
$ \begin{align}
& \Rightarrow \text{ }{{\left( AF \right)}^{2}}+{{\left( FG \right)}^{2}}={{\left( AG \right)}^{2}} \\
& \Rightarrow \text{ }{{\left( a \right)}^{2}}+{{\left( \sqrt{2}a \right)}^{2}}={{\left( AG \right)}^{2}} \\
& \Rightarrow \text{ }{{\left( AG \right)}^{2}}={{a}^{2}}+2{{a}^{2}} \\
& \Rightarrow \text{ }{{\left( AG \right)}^{2}}=3{{a}^{2}} \\
& \Rightarrow \text{ }AG=\sqrt{3}a \\
\end{align} $
As $ a=10cm $ , by substituting it in the above obtained answer, we get
$ \begin{align}
& \Rightarrow \text{ }AG=\sqrt{3}a \\
& \Rightarrow \text{ }AG=\sqrt{3}\times 10 \\
& \Rightarrow \text{ }AG=10\sqrt{3}cms \\
\end{align} $
So, we get that the length of the diagonal of the cube is equal to $ 10\sqrt{3}cms $ .
So, the length of the longest rod that can fit in a cube of side 10cms is $ 10\sqrt{3}cms $ .
So, the correct answer is “Option C”.
Note: There is a possibility of one making a mistake by considering the diagonal of a face of the cube as the diagonal of the cube, then the answer obtained will be $ 10\sqrt{2}cms $ , but it is wrong as the longest distance between two points in a cube is the diagonal between opposite edges of the cube not between that of the edges of the face of the cube.
Complete step-by-step answer:
We were given that the length of the side of the cube is 10cm.
We need to find the length of the longest rod that has to be fitted in the cube. As the rod has to touch the cube at two points, it means we have to find the longest distance between any two points on the sides of the cube.
Longest distance between two points of the cube is when it is the diagonal of the cube.
![seo images](https://www.vedantu.com/question-sets/071562d5-841f-4d3b-849a-05da2f8ade6f7188082198059071595.png)
Let the length of the side of the cube be $ a=10cm $ .
The longest distance is the diagonal of the cube which is AG from the above figure.
First, let us find the length of the side AF which is the diagonal of the face ABFE.
As ABFE is the face of a cube, it is square with side length $ a=10cm $ . Then $ \Delta ABF $ is a right-angled triangle.
So, by using the Pythagoras theorem, which is
If a, b are adjacent sides of a right-angled triangle and c is the hypotenuse, then
$ {{a}^{2}}+{{b}^{2}}={{c}^{2}} $ .
So, by applying the Pythagoras Theorem on the $ \Delta ABF $ , we get
$ \begin{align}
& \Rightarrow \text{ }{{\left( AB \right)}^{2}}+{{\left( BF \right)}^{2}}={{\left( AF \right)}^{2}} \\
& \Rightarrow \text{ }{{\left( a \right)}^{2}}+{{\left( a \right)}^{2}}={{\left( AF \right)}^{2}} \\
& \Rightarrow \text{ }{{\left( AF \right)}^{2}}=2{{a}^{2}} \\
& \Rightarrow \text{ }AF=\sqrt{2}a \\
\end{align} $
Now consider the $ \Delta AFG $ , where $ AF=\sqrt{2}a $ , $ FG=a $ as it is a side of the cube and $ \angle F={{90}^{\circ }} $ .
So, by applying the Pythagoras Theorem again on the $ \Delta ABF $ , we get
$ \begin{align}
& \Rightarrow \text{ }{{\left( AF \right)}^{2}}+{{\left( FG \right)}^{2}}={{\left( AG \right)}^{2}} \\
& \Rightarrow \text{ }{{\left( a \right)}^{2}}+{{\left( \sqrt{2}a \right)}^{2}}={{\left( AG \right)}^{2}} \\
& \Rightarrow \text{ }{{\left( AG \right)}^{2}}={{a}^{2}}+2{{a}^{2}} \\
& \Rightarrow \text{ }{{\left( AG \right)}^{2}}=3{{a}^{2}} \\
& \Rightarrow \text{ }AG=\sqrt{3}a \\
\end{align} $
As $ a=10cm $ , by substituting it in the above obtained answer, we get
$ \begin{align}
& \Rightarrow \text{ }AG=\sqrt{3}a \\
& \Rightarrow \text{ }AG=\sqrt{3}\times 10 \\
& \Rightarrow \text{ }AG=10\sqrt{3}cms \\
\end{align} $
So, we get that the length of the diagonal of the cube is equal to $ 10\sqrt{3}cms $ .
So, the length of the longest rod that can fit in a cube of side 10cms is $ 10\sqrt{3}cms $ .
So, the correct answer is “Option C”.
Note: There is a possibility of one making a mistake by considering the diagonal of a face of the cube as the diagonal of the cube, then the answer obtained will be $ 10\sqrt{2}cms $ , but it is wrong as the longest distance between two points in a cube is the diagonal between opposite edges of the cube not between that of the edges of the face of the cube.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)